I'm posting one puzzle, riddle, math, or statistical problem a day. Try to answer each one and post your answers in the comments section. I'll post the answer the next day. Even if you have the same answer as someone else, feel free to put up your answer, too!

Tuesday, July 31, 2007

What's Wrong With That?

The following is what seems to be a mathematical proof that ten equals 9.999999.... What's wrong with it?

11 comments:

The problem comes when you multiply by 10 and subtract. Since 9.999999... is not a rational number (that is, you can't express it in closed form with a fraction, like you could with, say 1.111111... = 10/9), we can only express it to some degree of accuracy (some number of significant digits, that is). Then, when you multiply by 10, the level of precision you chose is moved to the next level. If you do it this way, the problem resolves itself. For example:

a = 9.99910a = 99.9910a - a = 89.9919a = 89.991a = 9.999

The same principle applies for any number of significant figures you choose.

I used the definition of geometric sums, a/(1-r). I still say that level of precision argument I pointed out is sort of the crux of the trick in your proof, but apparently 9.99999.. = 10 in our decimal system for some reason. Good post, Mike.

Abe's response is the correct. The principal difference is the type of number: rationale or non-rationale. You can't mix orange and lemons :):):)When we say 1/3=0.333333..., that is an non-rationale aproximation. The correct is 1/3. The difference is minimum (correct) but there is a difference.So, when you're talk about "aproximation" the equations are correct: "The rationale approximation of the non-rationale 0.999999... is 1"

we can prove that 9.99999...... is a rational number and all other numbers in this format (repeated form) are rational numbers a = 9.999999...10a = 99.999999...10a - a = 909a = 90a = 10 by Zafarullah khan

11 comments:

The problem comes when you multiply by 10 and subtract. Since 9.999999... is not a rational number (that is, you can't express it in closed form with a fraction, like you could with, say 1.111111... = 10/9), we can only express it to some degree of accuracy (some number of significant digits, that is). Then, when you multiply by 10, the level of precision you chose is moved to the next level. If you do it this way, the problem resolves itself. For example:

a = 9.99910a = 99.9910a - a = 89.9919a = 89.991a = 9.999

The same principle applies for any number of significant figures you choose.

I used the definition of geometric sums, a/(1-r). I still say that level of precision argument I pointed out is sort of the crux of the trick in your proof, but apparently 9.99999.. = 10 in our decimal system for some reason. Good post, Mike.

Abe's response is the correct. The principal difference is the type of number: rationale or non-rationale. You can't mix orange and lemons :):):)When we say 1/3=0.333333..., that is an non-rationale aproximation. The correct is 1/3. The difference is minimum (correct) but there is a difference.So, when you're talk about "aproximation" the equations are correct: "The rationale approximation of the non-rationale 0.999999... is 1"

we can prove that 9.99999...... is a rational number and all other numbers in this format (repeated form) are rational numbers a = 9.999999...10a = 99.999999...10a - a = 909a = 90a = 10 by Zafarullah khan